Question

How much heat is evolved when 1255 grams of water condenses to a liquid at 100C?

Answer 1

Let's first convert grams of water to moles of water (molar mass of water is $18\,g/mol$):

$$1255\,g\,H_2O \cdot \left(\frac{1\,mol}{18.0\,g}\right) = 69.7\,mol\,H_2O.$$

The amount of heat released when $1\,mol$ of vapor condenses is called a molar heat of condensation ($\Delta H_{cond}$):

$$\Delta H_{vap} = -\Delta H_{cond}.$$

The molar heat of vaporization of water is $\Delta H_{vap} = 40.7\,kJ/mol$.

Then, $\Delta H_{cond} = -40.7\,kJ/mol$ and the next step is a conversion from moles of water to $\Delta H$, multiplying by the $\Delta H_{cond}$:

$$\Delta H = 69.7\,mol\,H_2O \cdot \frac{-40.7\,kJ}{1\,mol\,H_2O} = -2837\,kJ.$$

The negative sign indicate that heat is given off.

Therefore, the process will release $2837\,kJ$ of heat.

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